Dimension of the algebraic closure of a continuum field of characteristic zero Let me start by saying that I have no idea in algebra/number theory/whatever, so, please, forgive my ignorance.
Let $\mathbb{F}$ be a field of characteristic zero and continuum cardinality, which means its algebraic closure is (isomorphic to) $\mathbb{C}$ (complex numbers), right? Now $\mathbb{C}$ is an $\mathbb{F}$-algebra, but what is the $\mathbb{F}$-dimension of $\mathbb{C}$? In other words, if I pick up a Hamel basis in $\mathbb{C}$ over $\mathbb{F}$, what would be the cardinality of that basis? I imagine that it can be infinite, but can it be uncountable?
I discovered about the existence of such fields $\mathbb{F}$ other than $\mathbb{R}$ (real numbers) here:
http://mathforum.org/kb/message.jspa?messageID=7414811
Thank you.
 A: Suppose  $F\subsetneq  C$ is an arbitrary extension of fields of any characteristic $p\geq 0$, with $C$ algebraically closed and  with $[ C:F]=n\lt \infty$.
Then Artin-Schreier's  amazing theorem says that the characteristic  $p$ is necesarily zero, that $n=2$ and that there exists $j\in C$ such that $j^2=-1$ and $C=F(j)$.
Moreover $F$ must be real closed, which intuitively means that it resembles $\mathbb R$ viewed as a field, stripped of its non-algebraic structures.   
Bibliography
This theorem is the very last theorem, numbered 11.14 page 654, of Jacobson's Basic Algebra II.
Be assured however that you don't have to read the preceding 653 pages to understand Artin-Schreier, and even less the 499 pages of Basic Algebra I...
Edit
In a comment the OP asks whether it is possible to find  a subfield $F\subset \mathbb C$ with $\mathbb C$ algebraic over $F$ (so that $\mathbb C$ is an algebraic closure of $F$) and satisfying   $\dim_F \mathbb C\gt\aleph_0$.
The answer is yes:
Let $(x_i)_{i\in I}$ be a transcendence basis of $\mathbb C$ over $\mathbb Q$, necessarily of cardinality $\vert I\vert=\mathfrak c=2^{\aleph_0}$.
The field $F=\mathbb Q(x_i\vert i\in I)$ is the required field:
Indeed, if one chooses for each $i$  a square root $\sqrt x_i\in \mathbb C$, the family $(\sqrt x_i)_{i\in I}$ is $F$-linearly independent of cardinality $\vert I\vert=\mathfrak c$, which immediately implies that $\dim_F \mathbb C\geq \mathfrak c$ and thus (since the opposite inequality is trivial)  $$\dim_F \mathbb C=\mathfrak c=2^{\aleph_0}\gt \aleph_0.$$ (The linear independence of the $\sqrt x_i$'s is proved in the same elementary way that  the square roots $\sqrt p$ ($p$ prime) are linear independent over $\mathbb Q $: see for example here)
